A parallel combination of three resistors take a currect of `7.5 A` from `30 V` supply. If the two resistor are `10 Omega` and `12 Omega`. find the third one.

Answer

`R_(3) = 15 Omega`

Explanation

`R_(P) = 30//7.5 = 4 Omega`
`(1)/(R_(P)) = (1)/(R_(1)) + (1)/(R_(2)) + (1)/(R_(3)) or (1)/(4) + (1)/(10) + (1)/(12) + (1)/(R_(3))`
or `(1)/(R_(1)) = (1)/(4) - ((1)/(10) + (1)/(12)) = (4)/(60) = (1)/(15)`
or `R_(3) = 15 Omega`